A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th
1 answer:
Answer:
a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
The reduction on this case is
And since the new total time would be given by
b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time
The original time for INT operations is calculated as:
For this part the only time that was changed is assumed the INT operations so then:
And then:
c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.
Explanation:
From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.
Part 1
For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
The reduction on this case is
And since the new total time would be given by
Part 2
For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time
The original time for INT operations is calculated as:
For this part the only time that was changed is assumed the INT operations so then:
And then:
And we can quantify the decrease using the relative change:
of reduction
Part 3
A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.
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Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
Answer:
Explanation:
Previous concepts
Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:
Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:
MO = H˙ O
Principle of Angular Impulse and Momentum
The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:
Solution to the problem
For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is ".
If we analyze the staritning point we see that the initial velocity can be founded like this:
And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:
And if we integrate the left part and we simplify the right part we have
And if we solve for we got:
Answer:
the temperature of the aluminum at this time is 456.25° C
Explanation:
Given that:
width w of the aluminium slab = 0.05 m
the initial temperature = 25° C
h = 100 W/m²
The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;
density ρ = 2702 kg/m³
thermal conductivity k = 231 W/m.K
Specific heat c = 1033 J/Kg.K
Let's first find the Biot Number Bi which can be expressed by the equation:
Bi = 0.0108
The time constant value is :
Considering Lumped capacitance analysis since value for Bi is less than 1
Then;
where;
which correlates with the change in the internal energy of the solid.
So;
The maximum value for the change in the internal energy of the solid is :
By equating the two previous equation together ; we have:
Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75
Thus;
So;
t = 1.386294361 × 697.79
t = 967.34 s
Finally; the temperature of Aluminium is determined as follows;
T - 600 = -575 × 0.25
T - 600 = -143.75
T = -143.75 + 600
T = 456.25° C
Hence; the temperature of the aluminum at this time is 456.25° C